Let $R$ be the region enclosed by the curves $y=\sqrt x$ and $y=\dfrac x3$. $y$ $x$ $ R$ ${y=\sqrt x}$ ${y=\dfrac x3}$ Region $R$ is the base of a solid whose cross sections perpendicular to the $x$ -axis are squares. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_0^1 \left(\sqrt x-\dfrac x3\right)^2\,dx$ (Choice B) B $\int_0^3 \left(\sqrt x-\dfrac x3\right)^2\,dx$ (Choice C) C $\int_0^9 \left(\sqrt x-\dfrac x3\right)^2\,dx$ (Choice D) D $\int_0^{27} \left(\sqrt x-\dfrac x3\right)^2\,dx$
Solution: As a first step, lets find the exact coordinates of the intersection points of the graphs. To do that, we need to solve the equation $\sqrt x=\dfrac x3$. The solutions are $x=0$ and $x=9$. $y$ $x$ $ R$ ${y=\sqrt x}$ ${y=\dfrac x3}$ $(9,3)$ $(0,0)$ Now, let's imagine the solid is made out of many thin slices. $y$ $x$ $(9,3)$ $(0,0)$ Each slice is a prism. Let the width of each slice be $dx$ and let the area of the prism's face, as a function of $x$, be $A(x)$. Then, the volume of each slice is $A(x)\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(x)\,dx$ What we now need is to figure out the expression of $A(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\sqrt x}$ ${y=\dfrac x3}$ $(0,0)$ $(9,3)$ $ s(x)$ $ s(x)$ $ dx$ $ A(x)$ The face of that slice is a square with side $s(x)$. For each value of $x$, the side $s(x)$ is equal to the difference between ${y=\sqrt x}$ and ${y=\dfrac x3}$. Now we can find an expression for the area of the face of the prism: $\begin{aligned} &\phantom{=}A(x) \\\\ &=[s(x)]^2 \\\\ &=\left({\sqrt x}-{\dfrac x3}\right)^2 \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=9$. So the interval of integration is $[0,9]$. Now we can express the definite integral in its entirety! $\int_0^9 \left(\sqrt x-\dfrac x3\right)^2\,dx$